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A2 + 4A + 40 = (A2 + 4A + 4) + 36 = (A + 2)2 + (6)2 = (A + 2  j6)(A + 2 + j 6) llO 2 T imeDomain A nalysis o f C ontinuousTime S ystems where c and (J are arbitrary constants to be determined from the initial conditions yo(O) = 2
a nd yo(O) = 16.78. Differentiation of Eq. (2.12a) yields
yo(t) = _ 2ce 2t cos (6t + (J) 6ce 2t sin (6t + Ii) (2.12b)
S etting t = 0 in Eqs. (2.12a) and (2.12b), and then substituting initial conditions, we
o btain
2 =ccosli
 16.78 =  2ccos Ii  6csin Ii
Solution of these two simultaneous equations in two unknowns ccos
ccos (J = 2 (J and csin csin (J =  3.463
Squaring and t hen adding the two sides of the above equations yields
2 = (2)2 + ( 3.464)2 = 16 ==> c = 4
c
Next, dividing (2.13b) by (2.13a); t hat is dividing csin
t an (J =  3 463
2
and
(J Therefore by ccos (J yields
(2.13a)
(2.13b) yields = t an 1 (3 2
463) = 1 yo(t) = 4 e 2t cos (6t
Figure B .llc shows the plot of yo(t). o (J (J C omputer E xample C 2.1
Find the r oots of polynomial >2  1) • + 4> + 40 a =[l 4 40]; r = r oots(a) o r 2.0000 + 6 .0000i
 2.0000  6.0000i 0 C omputer E xample C 2.2
For an LTIC system specified by the differential equation
(D2 ( b) ( e) y O=dsolve('D2y+4*Dy+3*y=0','y(0)=3','Dy(0)=7','t')
yO = 2 *exp(3*t)+exp(t)
y O=dsolve('D2y+4*Dy+4*y=O' , 'y(O)=3', ' Dy(O)=7', ' t')
yO = 3 *exp(2*t)exp(2t)*t
y O=dsolve('D2y+4*Dy+40*y=O' , 'y(O)=3' , 'Dy(0)=7', ' t')
yO = 3 *exp(2*t)*eos(6*t)1/6*exp(2t)*sin(6*t) 0 6. E xercise E 2.1
F ind t he z eroinput r esponse o f a n LTIC s ystem described by (D+ 5)y(t) = f (t) i f t he i nitial
condition is y(O) = 5.
Answer: yo(t) = 5 e 5t
t:::: 0 'V 6. E xercise E 2.2
Solve (D2 + 2D) yo(t) = 0 i f Yo(O) = 1 a nd !io(O) = 4. Hint: T he c haracteristic roots a re 0 a nd  2.
Answer: yo(t) = 3  2e 2t t :::: 0 'V S ystem R esponse t o I nternal C onditions: Z eroInput R esponse 111 Practical Initial Conditions and the meaning of 0  and 0 +
I n E xample 2.1 t he i nitial c onditions YolO) a nd yolO) were supplied. I n p racticl\l
p roblems, we m ust d erive such conditions from t he p hysical s ituation. F or instance,
i n a n R LC c ircuit, we m ay b e g iven t he c onditions, s uch as i nitial c apacitor v oltages, a nd i nitial i nductor c urrents, e tc. F rom t his i nformation, we need t o d erive
yolO), yolO), . .. for t he d esired variable a s d emonstrated i n t he n ext e xample.
I n m uch o f o ur d iscussion, t he i nput is a ssumed t o s tart a t t = 0, u nless o therwise mentioned. Hence, t = 0 is t he r eference p oint. T he c onditions i mmediately
b efore t = 0 ( just b efore t he i nput is applied) a re t he c onditions a t t = 0 , a nd
t hose i mmediately a fter t = 0 ( just a fter t he i nput is applied) a re t he c onditions a t
t = 0 + ( compare t his w ith t he h istorical t ime f rame B .C. a nd A .D.). I n p ractice,
we a re likely t o k now t he i nitial c onditions a t t = 0  rather t han a t t = 0+. T he
t wo s ets o f conditions a re g enerally different, a lthough i n s ome cases t hey m ay b e
i dentical.
W e a re d ealing w ith t he t otal r esponse y (t), w hich c onsists o f two c omponents;
t he z eroinput...
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 Spring '13
 Bayliss
 Signal Processing, The Land

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